NSWMaths Extension 1Syllabus dot point

How do we use vectors to prove standard geometric results in the plane?

Use vector methods to prove geometric properties, including parallelism, perpendicularity, midpoint and ratio division

A focused answer to the HSC Maths Extension 1 dot point on geometric proofs using vectors. The standard techniques for showing two lines are parallel or perpendicular, the midpoint formula, the section formula, and complete worked proofs.

Generated by Claude OpusReviewed by Better Tuition Academy8 min answerUpdated 2026-05-20

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What this dot point is asking

NESA wants you to use the algebra of vectors to prove geometric properties: showing two lines are parallel, perpendicular, of equal length, share a midpoint, divide a segment in a given ratio, or that a figure is a parallelogram or rectangle.

The answer

Position vectors and triangle setup

Choose an origin $O$ . Each point $P$ has a position vector $\mathbf{p} = \mathbf{OP}$ .

For three points $A$ , $B$ , $C$ with position vectors $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ :

\mathbf{AB} = \mathbf{b} - \mathbf{a}, \qquad \mathbf{AC} = \mathbf{c} - \mathbf{a}.

This is the basic vocabulary for any geometric proof.

Parallelism

$\mathbf{u}$ and $\mathbf{v}$ are parallel iff $\mathbf{v} = \lambda \mathbf{u}$ for some scalar $\lambda$ . If $\lambda > 0$ , same direction; if $\lambda < 0$ , opposite direction. Equality of vectors ( $\mathbf{u} = \mathbf{v}$ ) is a stronger statement (parallel and equal length).

Perpendicularity

$\mathbf{u}$ and $\mathbf{v}$ are perpendicular iff $\mathbf{u} \cdot \mathbf{v} = 0$ .

Midpoint

The midpoint of $AB$ has position vector

\mathbf{M} = \frac{\mathbf{a} + \mathbf{b}}{2}.

This is the average of the two endpoints.

Section formula

The point dividing $AB$ internally in the ratio $m : n$ (so $\frac{AP}{PB} = \frac{m}{n}$ ) has position vector

\mathbf{P} = \frac{n \mathbf{a} + m \mathbf{b}}{m + n}.

For external division (the point is on the extension beyond $B$ ), use $\mathbf{P} = \frac{-n \mathbf{a} + m \mathbf{b}}{m - n}$ (with care over sign).

Parallelogram criterion

$ABCD$ is a parallelogram iff $\mathbf{AB} = \mathbf{DC}$ (opposite sides equal and parallel). Equivalently, the diagonals bisect each other: midpoint of $AC$ equals midpoint of $BD$ .

Rectangle criterion

A parallelogram is a rectangle iff one of its angles is right, i.e., $\mathbf{AB} \cdot \mathbf{AD} = 0$ .

Standard tactic

To prove a geometric statement:

Assign position vectors to all labelled points.
Express the relevant displacement vectors.
Compute (algebraically) the required relation: equality, dot product, scalar multiple, etc.
Conclude the geometric statement.

Avoid coordinates when the vector form is cleaner.

Worked example

Midpoints of a triangle's sides form a smaller triangle

Triangle $ABC$ . Let $P$ , $Q$ , $R$ be the midpoints of $BC$ , $CA$ , $AB$ . Show that $\mathbf{QR}$ is parallel to $BC$ and half its length.

Position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ for $A, B, C$ .

$P = \frac{\mathbf{b} + \mathbf{c}}{2}$ , $Q = \frac{\mathbf{c} + \mathbf{a}}{2}$ , $R = \frac{\mathbf{a} + \mathbf{b}}{2}$ .

$\mathbf{QR} = R - Q = \frac{\mathbf{a} + \mathbf{b}}{2} - \frac{\mathbf{c} + \mathbf{a}}{2} = \frac{\mathbf{b} - \mathbf{c}}{2}$ .

$\mathbf{BC} = \mathbf{c} - \mathbf{b}$ , so $\mathbf{QR} = -\frac{1}{2} \mathbf{BC}$ .

This means $\mathbf{QR}$ is parallel to $\mathbf{BC}$ (anti-parallel, but parallel as a line) and half the magnitude. Quod erat demonstrandum.

Diagonals of a parallelogram bisect each other

$ABCD$ is a parallelogram, so $\mathbf{AB} = \mathbf{DC}$ , that is $\mathbf{b} - \mathbf{a} = \mathbf{c} - \mathbf{d}$ , equivalently $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$ .

Midpoint of $AC$ : $\frac{\mathbf{a} + \mathbf{c}}{2}$ .

Midpoint of $BD$ : $\frac{\mathbf{b} + \mathbf{d}}{2}$ .

Since $\mathbf{a} + \mathbf{c} = \mathbf{b} + \mathbf{d}$ , the midpoints coincide. The diagonals bisect each other.

Cosine rule from vectors

$|\mathbf{a} - \mathbf{b}|^2 = (\mathbf{a} - \mathbf{b}) \cdot (\mathbf{a} - \mathbf{b}) = |\mathbf{a}|^2 - 2 \mathbf{a} \cdot \mathbf{b} + |\mathbf{b}|^2$ .

If $\theta$ is the angle between $\mathbf{a}$ and $\mathbf{b}$ , $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| |\mathbf{b}| \cos \theta$ , so

$|\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2 |\mathbf{a}| |\mathbf{b}| \cos \theta$ .

This is the cosine rule for a triangle with sides $|\mathbf{a}|$ , $|\mathbf{b}|$ and included angle $\theta$ .

Section in given ratio

$P$ divides $AB$ in the ratio $1 : 3$ . Find $P$ in terms of $\mathbf{a}$ and $\mathbf{b}$ .

$P = \frac{3 \mathbf{a} + 1 \mathbf{b}}{4} = \frac{3 \mathbf{a} + \mathbf{b}}{4}$ .

Perpendicular diagonals imply a rhombus

Show that if a parallelogram has perpendicular diagonals, it is a rhombus.

Let $ABCD$ be a parallelogram with $\mathbf{AB} = \mathbf{u}$ and $\mathbf{AD} = \mathbf{v}$ . Then $\mathbf{AC} = \mathbf{u} + \mathbf{v}$ and $\mathbf{BD} = \mathbf{v} - \mathbf{u}$ .

If diagonals are perpendicular, $(\mathbf{u} + \mathbf{v}) \cdot (\mathbf{v} - \mathbf{u}) = 0$ , which expands to $\mathbf{u} \cdot \mathbf{v} - |\mathbf{u}|^2 + |\mathbf{v}|^2 - \mathbf{u} \cdot \mathbf{v} = |\mathbf{v}|^2 - |\mathbf{u}|^2 = 0$ .

So $|\mathbf{u}| = |\mathbf{v}|$ : adjacent sides are equal in length, which is the rhombus condition.

Past exam questions, worked

Real questions from past NESA papers on this dot point, with our answer explainer.

2022 HSC Q265 marks

ABCD

is a quadrilateral. Let

P

Q

R

S

be the midpoints of

AB

BC

CD

DA

respectively. Use vectors to prove that

PQRS

is a parallelogram.

Show worked answer →

Let the position vectors of $A$ , $B$ , $C$ , $D$ be $\mathbf{a}$ , $\mathbf{b}$ , $\mathbf{c}$ , $\mathbf{d}$ .

Midpoints: $P = \frac{\mathbf{a} + \mathbf{b}}{2}$ , $Q = \frac{\mathbf{b} + \mathbf{c}}{2}$ , $R = \frac{\mathbf{c} + \mathbf{d}}{2}$ , $S = \frac{\mathbf{d} + \mathbf{a}}{2}$ .

Vector $\mathbf{PQ} = Q - P = \frac{\mathbf{b} + \mathbf{c}}{2} - \frac{\mathbf{a} + \mathbf{b}}{2} = \frac{\mathbf{c} - \mathbf{a}}{2}$ .

Vector $\mathbf{SR} = R - S = \frac{\mathbf{c} + \mathbf{d}}{2} - \frac{\mathbf{d} + \mathbf{a}}{2} = \frac{\mathbf{c} - \mathbf{a}}{2}$ .

So $\mathbf{PQ} = \mathbf{SR}$ , which means $PQ$ is parallel to $SR$ and equal in length.

Therefore $PQRS$ is a parallelogram.

Markers reward labelling position vectors of $A, B, C, D$ , computing each midpoint, computing both diagonals of the candidate parallelogram, and observing equality.